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#3280
by
deltaMass
on 14 May, 2015 02:35
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I'm only asking so as to help out with the maths. It doesn't mean I believe a word about its physicality.
For example: if the expression I'm trying to get verified is true, then you can get your blow-up condition. Other conditional relations produce different results (x0 < x1 < x2 or x1 < x2 < x0: x=lambda).
Then there's also the case that it blows up exactly when the numerator is also zero. As you know, 0/0 is an indeterminate quantity.
I find a nicer way to write Shawyer's Df is like this (again using x=lambda)
Df = x0*(x2 - x1) / (x1*x2 - x02)
You'll notice that Df=0 when x1 or x2 -> infinity (proof available on request but it's dead simple)
Someone said that Df varied between 0 and 1, and should =1 when x1 or x2 -> inf. It doesn't.
And obviously Df=0 when x1 = x2. So it can be zero for two separate reasons.
Doh!
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#3281
by
deltaMass
on 14 May, 2015 03:17
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I won't bother reading all that. From equation (5) of the IAC paper it's clear that any of the wavelengths in the waveguide are greater than the free space value. Or lambda1 > lambda0 and lambda2 > lambda0
Therefore your blow-up condition is impossible.
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#3282
by
TheTraveller
on 14 May, 2015 03:18
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....
As defined in the attachment dominant circular waveguide cutoff wavelength is 1.7 x diameter....
1.7 x diameter...
Yes of course, but what diameter? A cylinder has only one diameter. It has constant diameter.
Shawyer's EM Drive is not a cylinder with constant diameter. Shawyer's EM Drive is a truncated cone.
A truncated cone has a variable diameter. The diameters are different at each end.
Shawyer's reference (Cullen) does not deal with truncated cones.
What diameter ?
Look at what I derived in http://forum.nasaspaceflight.com/index.php?topic=36313.msg1373898#msg1373898:
...choosing the cutOffWavelength to be TE110,
Look at where the factor of 1.7 comes from: it is Pi divided by X'11
cOW = gmD *(Pi/1.84118378134065)
= gmD *1.7062895542683174
1.7 times the Geometric Mean of the Diameters
cOW = cutOffWavelength
gmD = Sqrt[bD * sD]
bD = big end diameter (m)
sD= small end diameter (m)
Shawyer only talks about the 2 ends of his cone and the wavelength of the Rf input.
Lambda0 is the wavelength of the input Rf signal in air or vacuum.
Lambda1 is the dominant cutoff wavelength for the bigger end of the cone.
Lambda2 is the dominant cutoff wavelength for the smaller end of the cone.
The focus on what is happening at each end of his cone cavity as the waves bounce off and transfer momentum, based on their individual group velocity, which varies as the dominant cutoff wavelength varies at each end. Higher group velocity at the big end and lower group velocity at the smaller end, delivering higher momentum xfer at the big end and lower momentum xfer at the smaller end.
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#3283
by
Rodal
on 14 May, 2015 03:21
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I'm only asking so as to help out with the maths. It doesn't mean I believe a word about its physicality.
For example: if the expression I'm trying to get verified is true, then you can get your blow-up condition. Other conditional relations produce different results (x0 < x1 < x2 or x1 < x2 < x0: x=lambda).
Then there's also the case that it blows up exactly when the numerator is also zero. As you know, 0/0 is an indeterminate quantity.
I find a nicer way to write Shawyer's Df is like this (again using x=lambda)
Df = x0*(x2 - x1) / (x1*x2 - x02)
You'll notice that Df=0 when x1 or x2 -> infinity (proof available on request but it's dead simple)
Someone said that Df varied between 0 and 1, and should =1 when x1 or x2 -> inf. It doesn't.
And obviously Df=0 when x1 = x2. So it can be zero for two separate reasons.
Doh!
Don't you have your Mathematica up and running ?
<<Df=0 when x1
or x2 -> infinity (proof available on request but it's dead simple) >> ??
Wrong. (You need AND) See:
In[16]:= Limit[x0*(x2 - x1)/(x1*x2 - x0^2), x1 -> Infinity]
Out[16]= -(x0/x2)In[17]:= Limit[x0*(x2 - x1)/(x1*x2 - x0^2), x2 -> Infinity]
Out[17]= x0/x1In[18]:= Limit[x0*(x2 - x1)/(x1*x2 - x0^2), x0 -> Infinity]
Out[18]= 0
In[19]:= Limit[x0*(x2 - x1)/(x1*x2 - x0^2), x1 -> 0]
Out[19]= -(x2/x0)
In[20]:= Limit[x0*(x2 - x1)/(x1*x2 - x0^2), x2 -> 0]
Out[20]= x1/x0
In[21]:= Limit[x0*(x2 - x1)/(x1*x2 - x0^2), x0 -> 0]
Out[21]= 0
In[22]:= Limit[Limit[x0*(x2 - x1)/(x1*x2 - x0^2), x1 -> Infinity],
x2 -> Infinity]
Out[22]= 0In[23]:= Limit[Limit[x0*(x2 - x1)/(x1*x2 - x0^2), x2 -> Infinity],
x1 -> Infinity]
Out[23]= 0In[24]:= Limit[Limit[x0*(x2 - x1)/(x1*x2 - x0^2), x1 -> 0],
x2 -> Infinity]
Out[24]= -\[Infinity]/x0
In[25]:= Limit[Limit[x0*(x2 - x1)/(x1*x2 - x0^2), x1 -> Infinity],
x2 -> 0]
Out[25]= x0 (-\[Infinity])
In[26]:= Limit[Limit[x0*(x2 - x1)/(x1*x2 - x0^2), x1 -> 0], x2 -> 0]
Out[26]= 0
In[27]:= Limit[Limit[x0*(x2 - x1)/(x1*x2 - x0^2), x2 -> 0], x1 -> 0]
Out[27]= 0
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#3284
by
deltaMass
on 14 May, 2015 03:28
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You missed my post above. In order for the group velocity in the waveguide to never exceed c (and it never does!) it follows from eqn(5) that all lambda's in the waveguide are greater than the lambda in free space.
Therefore 1 - (lambda02/(lambda1*lambda2)) is always positive definite.
Therefore blow-up is impossible
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#3285
by
Rodal
on 14 May, 2015 03:37
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You missed my post above. In order for the group velocity in the waveguide to never exceed c (and it never does!) it follows from eqn(5) that all lambda's in the waveguide are greater than the lambda in free space.
Therefore 1 - (lambda02/(lambda1*lambda2)) is always positive definite.
Therefore blow-up is impossible
I agree with you that the lambda in free space should be smaller than in the waveguide in principle.
I will carefully double check tomorrow his definitions and my derivation
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#3286
by
TheTraveller
on 14 May, 2015 03:52
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You missed my post above. In order for the group velocity in the waveguide to never exceed c (and it never does!) it follows from eqn(5) that all lambda's in the waveguide are greater than the lambda in free space.
Therefore 1 - (lambda02/(lambda1*lambda2)) is always positive definite.
Therefore blow-up is impossible
I agree with you in principle. I have to double check where he puts the cutoff wavelength...
All wavelengths inside the conic cavity are bigger than outside and the resultant group velocities are less than the velocity outside.
The guide wavelength and resultant group velocity constantly varies, driven by the constantly varying diameter of the conic section the wave is passing through.
Also the edges of the wave fronts are at right angles to the cone sides due to being spherical wave fronts as if they originated from and are returning to the vertex of the cone.
If however the conic cavity end plates are flat and not spherically matching the spherical wave fronts bouncing off them, well you may be pushing s##t up hill before things are working well inside the cavity.
I'm actually amazed a flat ended conic cavity can generate thrust.
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#3287
by
Rodal
on 14 May, 2015 04:05
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You missed my post above. In order for the group velocity in the waveguide to never exceed c (and it never does!) it follows from eqn(5) that all lambda's in the waveguide are greater than the lambda in free space.
Therefore 1 - (lambda02/(lambda1*lambda2)) is always positive definite.
Therefore blow-up is impossible
I agree with you in principle. I have to double check where he puts the cutoff wavelength...
All wavelengths inside the conic cavity are bigger than outside and the resultant group velocities are less than the velocity outside.
The guide wavelength and resultant group velocity constantly varies, driven by the constantly varying diameter of the conic section the wave is passing through.
Also the edges of the wave fronts are at right angles to the cone sides due to being spherical wave fronts as if they originated from and are returning to the vertex of the cone.
If however the conic cavity end plates are flat and not spherically matching the spherical wave fronts bouncing off them, well you may be pushing s##t up hill before things are working well inside the cavity.
So Shawyer does not have a cut-off wavelength appearing anywhere on his Design Factor ?
If so, the Design Factor should be trivial, much easier to calculate...
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#3288
by
TheTraveller
on 14 May, 2015 04:14
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You missed my post above. In order for the group velocity in the waveguide to never exceed c (and it never does!) it follows from eqn(5) that all lambda's in the waveguide are greater than the lambda in free space.
Therefore 1 - (lambda02/(lambda1*lambda2)) is always positive definite.
Therefore blow-up is impossible
I agree with you in principle. I have to double check where he puts the cutoff wavelength...
All wavelengths inside the conic cavity are bigger than outside and the resultant group velocities are less than the velocity outside.
The guide wavelength and resultant group velocity constantly varies, driven by the constantly varying diameter of the conic section the wave is passing through.
Also the edges of the wave fronts are at right angles to the cone sides due to being spherical wave fronts as if they originated from and are returning to the vertex of the cone.
If however the conic cavity end plates are flat and not spherically matching the spherical wave fronts bouncing off them, well you may be pushing s##t up hill before things are working well inside the cavity.
So Shawyer does not have a cut-off wavelength appearing anywhere on his Design Factor ?
He uses what the industry uses. Guide wavelength as in the attached. It is related to cutoff wavelength as per the attached equation
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#3289
by
deltaMass
on 14 May, 2015 04:16
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Df = 0 when lambda1 = lambda2
Df = 1 when min(lambda1, lambda2) = lambda0 (free space condition)
So it looks like the right Df
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#3290
by
Rodal
on 14 May, 2015 04:17
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You missed my post above. In order for the group velocity in the waveguide to never exceed c (and it never does!) it follows from eqn(5) that all lambda's in the waveguide are greater than the lambda in free space.
Therefore 1 - (lambda02/(lambda1*lambda2)) is always positive definite.
Therefore blow-up is impossible
Back to basics. How does Shawyer define
lambda0 ? Does he define lambda0 = loFreeSpaceWavelength = c/rfFrequency ?
How does Shawyer define lambda 1 and lambda 2 ?
He does define lambda 1 not equal to lambda2, so they are different, we know that
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#3291
by
Rodal
on 14 May, 2015 04:21
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You missed my post above. In order for the group velocity in the waveguide to never exceed c (and it never does!) it follows from eqn(5) that all lambda's in the waveguide are greater than the lambda in free space.
Therefore 1 - (lambda02/(lambda1*lambda2)) is always positive definite.
Therefore blow-up is impossible
I agree with you in principle. I have to double check where he puts the cutoff wavelength...
All wavelengths inside the conic cavity are bigger than outside and the resultant group velocities are less than the velocity outside.
The guide wavelength and resultant group velocity constantly varies, driven by the constantly varying diameter of the conic section the wave is passing through.
Also the edges of the wave fronts are at right angles to the cone sides due to being spherical wave fronts as if they originated from and are returning to the vertex of the cone.
If however the conic cavity end plates are flat and not spherically matching the spherical wave fronts bouncing off them, well you may be pushing s##t up hill before things are working well inside the cavity.
So Shawyer does not have a cut-off wavelength appearing anywhere on his Design Factor ?
He uses what the industry uses. Guide wavelength as in the attached. It is related to cutoff wavelength as per the attached equation
No, he doesn't use the standard definition because according to that formula there is only one waveguide wavelength and he is defining two of them.
Same question that deltaMass was asking
How does Shawyer define lambdag 1 and lambdag 2 ?
He does define lambdag1 not equal to lambdag2,
so they are different, we know thatwhat is the difference?
In the formula you show the only things that appear are the free space wavelength and the cut off wavelength
But there is only one free space wavelength
There is only one cutoff wavelength
SO WHAT IS THE DIFFERENCE between lambdag1 and lambdag2
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#3292
by
TheTraveller
on 14 May, 2015 04:24
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Df = 0 when lambda1 = lambda2
Df = 1 when min(lambda1, lambda2) = lambda0 (free space condition)
So it looks like the right Df
Shawyers definitions for Df = 0 and Df = 1 are attached.
Your Df = 0 assumption matches Shawyer but for Df = 1, small end plate lambda2 must be at cutoff and big end plate lambda1 close to lambda0.
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#3293
by
deltaMass
on 14 May, 2015 04:25
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Rodal, you seem confused. Lambda0 is the free space wavelength = c/f. Equation (5) tells you the rest.
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#3294
by
deltaMass
on 14 May, 2015 04:26
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The Traveller: I know, and what I said doesn't contradict you, except that cut-off isn't necessary.
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#3295
by
Rodal
on 14 May, 2015 04:28
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Rodal, you seem confused. Lambda0 is the free space wavelength = c/f. Equation (5) tells you the rest.
that's the same definition of free space wavelength I gave you
What is lambdag 1 and lambdag 2 ?
Please write them down in your message
Thanks
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#3296
by
deltaMass
on 14 May, 2015 04:32
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They are what TheTraveller described and what's in eqn (5). To whit:
vgN / c = lambda0 / lambdagN
for N=1,2,3,4,5,6...whatever
...assuming the relative permeability and relative permittivity both =1
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#3297
by
TheTraveller
on 14 May, 2015 04:34
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You missed my post above. In order for the group velocity in the waveguide to never exceed c (and it never does!) it follows from eqn(5) that all lambda's in the waveguide are greater than the lambda in free space.
Therefore 1 - (lambda02/(lambda1*lambda2)) is always positive definite.
Therefore blow-up is impossible
I agree with you in principle. I have to double check where he puts the cutoff wavelength...
All wavelengths inside the conic cavity are bigger than outside and the resultant group velocities are less than the velocity outside.
The guide wavelength and resultant group velocity constantly varies, driven by the constantly varying diameter of the conic section the wave is passing through.
Also the edges of the wave fronts are at right angles to the cone sides due to being spherical wave fronts as if they originated from and are returning to the vertex of the cone.
If however the conic cavity end plates are flat and not spherically matching the spherical wave fronts bouncing off them, well you may be pushing s##t up hill before things are working well inside the cavity.
So Shawyer does not have a cut-off wavelength appearing anywhere on his Design Factor ?
He uses what the industry uses. Guide wavelength as in the attached. It is related to cutoff wavelength as per the attached equation
No, he doesn't use the standard definition because according to that formula there is only one waveguide wavelength and he is defining two of them.
Same question that deltaMass was asking
How does Shawyer define lambdag 1 and lambdag 2 ?
He does define lambdag1 not equal to lambdag2, so they are different, we know that
what is the difference?
In the formula you show the only things that appear are the free space wavelength and the cut off wavelength
But there is only one free space wavelength
There is only one cutoff wavelength
SO WHAT IS THE DIFFERENCE between lambdag1 and lambdag2
The cutoff wavelength is defined for EACH end, based on it's diameter * 1.71. See attachment 2.
Lambda1 = guide wavelength equation from attachment 1 below using big end cutoff wavelength.
Lambda2 = guide wavelength equation from attachment 1 below using small end cutoff wavelength.
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#3298
by
Rodal
on 14 May, 2015 04:37
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....
The cutoff wavelength is defined for EACH end, based on it's diameter * 1.71. See attachment 2.
Lambda1 = guide wavelength equation from attachment 1 below using big end cutoff wavelength.
Lambda2 = guide wavelength equation from attachment 1 below using small end cutoff wavelength.
OK
Thanks for the answer
Very clear answer
That's NOT what deltaMass had written (there is no N=1,2,3,4, in equation 5)
That's exactly the same factor of 1.7 I derived, based on TE110Now I'm going to sleep.
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#3299
by
deltaMass
on 14 May, 2015 04:39
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Notice that the game may change if sqrt(mur epsr) < 1. Usually it's ~=1, or a bit more , and that's the assumption I'm making with my inequalities games